Let $R$ be an ordered field with the Nested Interval Property: If $I_1,I_2,...,I_n,...$ be a collection of nested closed intervals, i.e $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ then $\bigcap_{1}^{\infty} I_i \neq \phi$ where $I_n=[a_n,b_n]$.
Now If $(b_n-a_n) \to 0$ as $n \to \infty$. Then $\bigcap_{1}^{\infty} I_i$ is singleton.
Let $s_1,s_2$ be two candidates. If $s_1 \neq s_2$ then WLOG let $s_2-s_1>0$. As $(b_n-a_n) \to 0$ there exists $n \in \mathbb{N}$ such that $b_n-a_n<s_1-s_2$ $\implies [a_n,b_n]$ cannot contain both $s_1$ and $s_2$, contradiction.
Hence $s_1=s_2.$
Is this proof correct? Then it means that any ordered field with the Nested Interval Property will have this property. Am I correct?